# Definition of compactly supported functions

Let $$X$$ be a topological space. Let $$f \in C(X)$$. Define the support of $$f$$. $$\operatorname{supp}(f) := \overline{\{x \in X: f(x) \neq 0\}}$$

I want to show that $$A:=\{f \in C(X)\mid \exists K \subseteq X \mathrm{\ compact \ }: \forall x \notin K: f(x) = 0\}$$

and $$B:=\{f \in C(X): \operatorname{supp}(f) \mathrm{\ is \ compact}\}$$

coincide. The inclusion $$B \subseteq A$$ is trivial.

Attempt: I managed to show that $$A \subseteq B$$ if $$X$$ is Hausdorff:

If $$f \in A$$, determine a compact set $$K$$ with $$f(x) = 0$$ for $$x \notin K$$. Then $$\{x \in X: f(x) \neq 0\} \subseteq K$$ and taking closures $$\operatorname{supp}(f) \subseteq \overline{K}$$ Since $$X$$ is Hausdorff, $$K$$ is closed so we get $$\operatorname{supp}(f) \subseteq K$$

Then $$\operatorname{supp}(f)$$ is a closed subset of a compact set and we can conclude.

Question: Does the inclusion remain valid without the Hausdorff assumption?

• This is a very good question in my opinion. – JustDroppedIn Jul 7 at 10:51

Not necessarily. You can take as an example a modification of the line with two zeros - in particular, let $$C$$ be an infinite set and let $$X=C\cup(0,1].$$ Define $$f:X\rightarrow\mathbb R$$ by $$f(x) = \begin{cases}0 & \text{if }x\in C\\x & \text{if }x\in (0,1].\end{cases}$$ Define a topology on $$X$$ by saying that a set $$U\subseteq X$$ is open if and only if $$f(U)$$ is open. This really does define a topology, though you should carefully check what happens with intersections of open sets.
Clearly, $$f$$ is continuous. Moreover, if $$c\in C$$, then $$\{c\}\cup (0,1]$$ is homeomorphic by $$f$$ to $$[0,1]$$ - meaning that $$f\in A$$ since this is a compact set containing all non-zero values of $$f$$. However, the support of $$f$$ is all of $$X$$, which is not compact because the cover $$\{\{c\} \cup (0,1] : c\in C\}$$ has no non-trivial subcovers (and hence no finite subcovers). Thus $$f\not\in B$$.